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Elden [556K]
3 years ago
14

1. Vector B is multiplied by the scalar 921.5. What angle does the new vector make with the x-axis?

Physics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

#1 is actually 33 because I’m assuming the angle doesn’t change just because the vector line gets longer

Explanation: I took this question on physics test and picked 49.5 and I got it wrong

adelina 88 [10]3 years ago
6 0
1. Is 49.5
2. Is 8.6
4. Is 6.6
Wait I’m not sure
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Horace has invented a unique pair of reading glasses that have two small light bulbs at the bottom wired in series, so that he c
ElenaW [278]

Answer:

The current drawn by Horace’s reading glasses is 0.8 A.

Explanation:  

Given that,

Resistance of each bulb, R = 2 ohms

Voltage of the system, V = 3.2 volts

These two bulbs are connected in series. The equivalent resistance will be 2 ohms +2 ohms = 4 ohms

Let I is the current drawn by Horace’s reading glasses. Using Ohm's law to find it such that :

V=IR\\\\I=\dfrac{V}{R}\\\\I=\dfrac{3.2}{4}\\\\I=0.8\ A

So, the current drawn by Horace’s reading glasses is 0.8 A.

3 0
4 years ago
All phase changes are .an example of a phase change is an ice cube melting
masha68 [24]

Answer:

a

Explanation:

a

8 0
3 years ago
Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its
kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2},

Why does this formula work?

By Faraday's Law of Induction, the EMF \epsilon induced in a coil (one loop) is equal to the rate of change in the magnetic flux \Phi through the coil.

\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t)).

Finding the average EMF in the coil is similar to finding the average velocity.

\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt.

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0).

Hence the equation

\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}.

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in \Phi(t) won't matter.

Apply this formula to this question. Note that \Phi, the magnetic flux through the coil, can be calculated with the equation

\Phi = B \cdot A \cdot N \; \sin{\theta}.

For this question,

  • B = \rm 16\; mT = 16\times 10^{-3}\; T is the strength of the magnetic field.
  • A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2} is the area of the coil.
  • N = 1 is the number of loops in the coil.
  • \theta is the angle between the field lines and the coil.
  • At \rm 0\;s, the field lines are parallel to the coil, \theta = 0^{\circ}.
  • At \rm 0.7\; s, the field lines are perpendicular to the coil, \displaystyle \theta = 90^{\circ}.

Initial flux: \Phi(0)= 0.

Final flux: \Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb.

Average EMF, which is the same as the average rate of change in flux:

\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V.

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4 years ago
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3 years ago
What type of force pulls in two opposite directions?
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Tensional forces which is associated with normal faults
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